invertible matrix

Let $R$ be a ring and $M$ an $m\times n$ matrix over $R$. $M$ is said to be left invertible if there is an $n\times m$ matrix such that $NM=I_n$, where $I_n$ is the $n\times n$ identity matrix. We call $N$ a left inverse of $M$. Similarly, $M$ is right invertible if there is an $n\times m$ matrix $P$, called a right inverse of $M$, such that $MP=I_m$, where $I_m$ is the $m\times m$ identity matrix. If $M$ is both left invertible and right invertible, we say that $M$ is invertible. If $R$ is an associative ring, and $M$ is invertible, then it has a unique left and a unique right inverse, and they are in fact equal, we call this matrix the inverse of $M$.

If $R$ is a division ring, then it can be shown that for any matrix $M$ over $R$, $M$ is left invertible iff it is invertible iff it is right invertible. In addition, when $M$ is invertible, it is a square matrix. Furthermore, $R$ is a field iff for any square matrix $M$ (over $R$), $M$ is invertible implies that $M^T$, its transpose, is invertible as well. Invertibility of matrices over a division ring can also be determined by quantities known as ranks and determinants. It can be shown that a matrix over a division ring is invertible iff its left row rank (or right column rank) is full iff its determinant is non-zero. For example, the $2\times 2$ matrix

over the Hamiltonian quaternions is not invertible, as its determinant $k-ji=0$. It is interesting to note that, however, its transpose

is invertible, whose determinant is $2k\ne 0$. The relationship between determinants and matrix invertibility can also be used to prove the following: preservation of matrix invertibility upon matrix transposition implies commutativity of division ring $D$. This can be done as follows: given any $a,b\in D$, the $2\times 2$ matrix

is not invertible because its determinant is $0$. Therefore, its transpose

is also not invertible, and its determinant is $0=ab-ba$, whence $D$ is a field.